Question 1048773
h+r = 60, and {{{V = pi*r^2h}}}  ===> {{{V = pi*r^2(60-r)}}}

===> V' = {{{pi*(2r(60-r)-r^2) = pi*(120r - 3r^2)}}}

Letting V' = 0, we get r = 0, 40.

Eliminate r = 0.  ===> Absolute extremum at r = 40.

Now V" = {{{pi*(120 - 6r)}}}  ===> V"(40) = {{{pi*(120 - 6*40) = -120 < 0)}}} 

===> absolute maximum when r = 40.

===> h = 60 - 40 = 20 cm.

===> Maximum volume is {{{V = pi*40^2*20 = 32000pi)}}} cm^3.